# These are the detailed Difficult Puzzle Grid Solutions :

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Below is the complete listing for all of the Difficult Puzzle Grid Solutions!

This page contains the DIFFICULT puzzle grid solutions to all of the PDF logic puzzle grids on our web site.

( You may consider this page a sort of cheat or crib sheet for our logic puzzles.)

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DIFFICULT :

## DIFFICULT :

1.

FIVE CLOWNS-SOLUTION

## FIVE CLOWNS-SOLUTION (show / hide)

Argo-Seals-130pm
Bargo-Elephants-330pm.
Largo-Balloons-215pm.
Margo-Tricycle-515pm.
Vargo--Monkeys-230pm

#### step-by-step:( show / hide)

• Lets look at the first clue "Vargo is scheduled to perform before the one who does tricks with elephants , but after Largo ."

• Lets decipher this clue as follows: Vargo does not do the tricks with elephants(nor does Largo), therefore
Eliminate grid squares [Vargo-Elephant, Largo-Elephant].

Now lets tackle the time elements mentioned.
'...Vargo scheduled before .....but after Largo', which means Vargo performs neither 1st(130pm), nor last (515pm).
Eliminate [Vargo-130pm, Vargo-515pm], and since Vargo is scheduled to perform 'before...elephants'
Eliminate [130pm-Elephants] and since Largo appears before at least both, eliminate [Largo-330pm, Largo-515pm].

• "Bargo is not scheduled to perform either last, nor first."

• Simply eliminate [Bargo-130pm, Bargo-515pm].

• "Margo has a wonderful tricycle high-wire riding act."

• Locate grid square [Margo-Tricycle] and click until the 'green box' appears, then make the following eliminations
[Margo-Balloons, Margo-Elephant, Margo-Monkey, Margo-Seals] as well as
[Argo-Tricycle, Bargo-Tricycle, Largo-Tricycle, Vargo-Tricycle].

• "The Seals are scheduled to go on before the monkeys who go on after the one who makes balloon animals-who is not Argo , who is not performing at 5:15pm."

• We will break this down into smaller fragments as follows (taking the last part first) :

We are given the clues 'Argo does not make balloon animals or perform at 515pm', so eliminate
[Argo-Balloons, Argo-515pm], and by doing so we reveal the name of the 515pm performer: Margo.

So after selecting [Margo-515pm] click until the 'green box appears, and make the following grid square eliminations
[Margo-130pm, Margo-215pm, Margo-230pm, Margo-330pm] and [Tricycle-130pm, Tricycle-215pm, Tricycle-230pm, Tricycle-330pm] and after selecting [Tricycle-515pm] click until the 'green box' appears there,
then proceed to make the eliminations:
[515pm-Balloons, 515pm-Elephants, 515pm-Monkey, 515pm-Seals]

Now for the first part of the clue: "...Seals ...before Monkeys..who go on after ..balloons", allows us to make the
following eliminations [Monkey-130pm, Monkey-215pm] because monkeys go on after 'BOTH' 'Seals & Balloons'.

• "Largo performs three hours before Margo , and will not be using real animals in his act."

• Since we now know 'Margo performs at 515 pm', Largo must appear at 215pm.
( Therefore Click on grid square [Largo-215pm] until the 'green box' appears), and make the
following eliminations: [Largo-130pm, Largo-250pm, Argo-215pm, Bargo-215pm, Vargo-215pm],

which reveals 'Argo's performance time as 130pm.'
( Click grid square [Argo-130pm] until the 'green box' appears), which leads to the eliminations
[Argo-230pm, Argo-330pm].

NOTE: You can also make the following elimination [Balloons-130pm] because we have already eliminated
[Argo-Balloons], which reveals the act performed at 130pm as 'Seals,

So after clicking and selecting the green box for [130pm-Seals], we make our eliminations
[215pm-Seals, 230pm-Seals,330pm-Seals], and since we know
'Argo performs at 130pm' and the 'Seal act is at 130pm' , it follows that :

"Argo performs the Seal act at 130pm",

therefore locate grid [130pm-Seals] and click until the green box appears,
after which we make our eliminations:

[Argo-Elephant, Argo-Monkey, Bargo-Seals,Largo-Seals,Vargo-Seals],
which then reveals another solution, namely, [Bargo-Elephant] and we know Bargo performs at either 230pm or 330pm,
So we can eliminate [Elephant-215pm], which reveals yet, still another solution [215pm-Balloons],
and more eliminations: [230pm-Balloons, 330pm-Balloons].
Now that we know that 'Largo performs at 215 pm' , we then know he MUST be the clown who uses the

'balloon animals', so select [Largo-215pm] while eliminating [Largo-Monkey],

and finishing our Clown-Act squares by selecting [Vargo-Monkey] and eliminating [Vargo-Balloons],

At this point it only remains to figure out which performer(of Bargo or Vargo) who appears at which time
(only 230pm & 330pm time slots remain).

Which brings us to this final clue :

• "Vargo is scheduled to appear at least one hour before one act and at least one hour , (but not two) after the first act."

• Which means Vargo's act appears at 230pm!

• Congratulations! Puzzle solved. To summarize:

Argo performs with the Seals at 130pm.
Bargo performs with the Elephants at 330pm.
Largo performs with the Balloons at 215pm.
Margo performs with the Tricycle at 515pm.
Vargo performs with the Monkeys at 230pm.
###### PDF PRINTABLE VERSION

2.

ON THE FARM-SOLUTION

## ON THE FARM-SOLUTION (show / hide)

Adam- Monday -8am-Milks bessie the goat.
Carl-Tuesday-7am-collects eggs.
Fran-Wednesday-5am-Mends the fences.
Nan-Thursday-9am -plant the corn.
Mike-Friday-6am-Market

#### step-by-step:( show / hide)

• Lets look at the first clue "Fran's mending fence day is neither Tuesday nor Thursday , and is completed earlier in the week than at least one other person ."

• The first part of this clue "...Fran ....mending fence day.....neither Tuesday nor Thursday" yields a solution [Fran-Mend]
and elimination grid squares [Fran-Milk, Fran-Market, Fran-Corn, Fran-Eggs] as well as[Fran-Tuesday, Mend-Tuesday, Fran-Thursday, Mend-Thursday] and [Adam-Mend, Carl-Mend, Nan-Mend, Mike-Mend]
Eliminate [Fran-Friday, Mend-Friday]since "... Fran's...mending...is...earlier in the week than at least one other person."

• "The one who goes to market goes later in the week than Carl , and earlier in the morning than Nan but, later in the morning than Fran. ."

• Eliminate [Carl-Market,Nan-Market, Fran-Market], since all three of these people are compared to the"... one who goes to market..."
Eliminate [Market-Monday]since "... market...later in the week than Carl" and eliminate [Carl-Friday].
Eliminate [Market-9am, Nan-5am] since "... market...earlier in the morning than Nan"
Eliminate [Market-5am, Fran-8am, Fran-9am, Mend-8am, Mend-9am] since "... market...later in the morning than Fran"

• "Adam does not plant the corn , or go to market his chore is completed, before 9am but not before 6am on his assigned day of Monday ."

• Eliminate [Adam-Market,Adam-Corn,], and note that leaves a solution for the person who goes to market, (by the process of elimination) must be Mike, so find square [Mike-Market] and click until a 'green box' appears, (which, of course) yields the eliminations [Mike-Milk, Mike-Corn, Mike-Eggs]

Returning to finish the clue for Adam we are given the times "...before 9am, but not before 6am, which results in the following eliminatons:[Adam-5am, Adam-6am, Adam-9am]. Finally to complete Adam's clue we have : "...his assigned day of Monday" so click square[Adam-Monday] until the 'green box' appears there.
*NOTE* : The completion of this clue(if done correctly) uncovers a solution for Fran in the grid : 'Fran-Wednesday' (and by association of Fran-Mend) 'Mend-Wednesday'.

So locate these two grid squares[Fran-Wednesday, Mend-Wednesday] and click until both are loaded with 'green boxes', (which allows us to make these eliminations:[Carl-Wed, Nan-Wed, Mike-Wed, Milk-Wed, Market-Wed, Corn-Wed, Eggs-Wed, 8-Wed, 9-Wed] )

• "No one performs a chore on Thursday at 7am ."

• Eliminate [7-Thursday]

• "Mike's chore is completed earlier in the day than at least 3 others ,but later in the week than at least three others ."

• NOTE: Mike's chore is 'Market' (so keep that in mind for the following eliminations)
Mike-Market is " ...earlier ...day than at least 3 others
Eliminate [Mike-7, Mike-8, Mike-9, Market-7, Market-8]
*IMPORTANT * : These eliminations uncover a solution- [Market-6] and by association (of Mike-market) [Mike-6], so both squares should be filled with 'green boxes' and make the eliminations:
[Mike-5 , Carl-6, Nan-6, Fran-6] as well as [Milk-6, Corn-6, Mend-6, Eggs-6, 6-Wed].
Mike-Market is " ...later ...week than at least 3 others
Eliminate [Mike-Tuesday, Market-Tuesday]

• "One man collects eggs, another man milks Bessie, the goat ."

• Eliminate [Nan-Eggs, Nan-Milk] because the key here is 'MAN', and since Nan is not a man, we eliminate those squares, which of course leads to the solution: [Nan-Corn], (which in turn eliminates : [Carl-Corn,Corn-5].

• "The milk is collected earlier in the week than the woman who plants the corn which is not planted before wednesday."

• The first part of this clue allows us to eliminate [Milk-Fri], so that we can now focus on the second part of the clue.
To begin with we now know this woman is Nan, thus we can make the eliminations [Nan-Tue, Corn-Tue] because "corn...is not..before Wednesday"
This means the only person left to perform chores on Tuesday is 'Carl' So we select
[Carl-Tue] and eliminate [Carl-Thu].

• "The eggs must be collected at 7 am. , but never on mondays or fridays."

• First locate [Eggs-7] and click until 'green', then make the eliminations:
[Eggs-5, Eggs-6, Eggs-8, Eggs-9] as well as [Milk-7, Market-7,Corn-7, Mend-7].
Which leads to the logical conclusions [Mend-5, Fran-5, 5-Wed] and the eliminations [Carl-5, Fran-7, 5-Tue, 5-Thu, 5-Fri, 7-Wed]
The last part of the clue states "...Eggs...collected....never on...mondays or fridays", which allows us to make the following logical eliminations:
[Eggs-Mon, Eggs-Fri]. This will yield another solution: [Milk-Monday], which also means [Adam-Milk], because Adam-Mon.
Now the resultant eliminations :[Adam-Eggs, Carl-Milk, Milk-Tue, Milk-Wed, Milk-Thu, Milk-Fri, Eggs-9]
Which, in turn, reveals still more solutions [Carl-Eggs, Eggs-Tue, Milk-8, Adam-8, Monday-8,] and their resulting eliminations:
[Adam-7, Carl-8, Nan-8,8-Tue, 8-Thu, 8-Fri, Eggs-Thu, Corn-8 ], and even more solutions:
[Corn-9, Nan-9, Carl-7, 7-Tue ]

• "All the other chores must be complete before Mike goes to the market."

• In other words, Mike( and Market) are completed last(Friday). Locate and click on [Mike-Fri, Market-Fri] until 'green boxes' are added to both.

• Congratulations! Puzzle solved. To summarize:
Adam- Monday -8am-Milks bessie the goat.
Carl-Tuesday-7am-collects eggs.
Fran-Wednesday-5am-Mends the fences.
Nan-Thursday-9am -plant the corn.
Mike-Friday-6am-Market
###### PDF PRINTABLE VERSION

3.

COSTUMES & CANDY-SOLUTION

## COSTUMES & CANDY- show / hide-

Bart-Pirate-12.
Gina-Witch-2.
Ginger-Nurse-8 .
Susie-Princess-10.
Sam-Ghoul-6.

#### step-by-step:( show / hide)

• This first clue "One boy dressed as a ghoul."

• Simply eliminate 'All' the girls ( Gina, Ginger, Susie) grid squares as follows:
[Gina-Ghoul, Ginger-Ghoul, Susie-Ghoul].

• This next clue : "The boy in the pirate costume received more candy than the other 2 boys."

• Again we can eliminate 'All' the girls grid squares that meet with the column Pirate
[Gina-Pirate, Ginger-Pirate, Susie-Pirate]. The only difference is we have an additional portion of the clue, namely
"...Boy... Pirate...received more candy than....", which means the pirate received more than at least 2 others, so eliminate the grid squares:
[Pirate-2, Pirate-4].

• This clue says : "The girl in the Witch costume received fewer pieces of candy than the other 2 girls.."

• Similar to the previous clue, we can eliminate "ALL" of the boys squares that intersect with the column Witch
[Bart-Witch, Brad-Witch, Sam-Witch]. The only difference is we have an additional portion of the clue, namely
"...Girl... Witch...received ...fewer than....", which means the Witch received less than at least 2 others, so eliminate the grid squares:
[Witch-10, Witch-12].

• This next clue "Brad (who was a vampire) received less candy than the other two boys , but more than Gina."

• Provides our first solution, so locate grid square [Brad-Vampire] and highlight that with a 'green-box', (we can also eliminate the remaining kid's names from the Vampire Column) :
Vampire - Bart, Gina, Ginger, Susie, and Sam. We can also clear some entries from Brad's row:
Brad - Pirate, Ghoul, Witch, Nurse, and Princess.

**NOTE** (we can also eliminate the Witch,Nurse, and Princess) from both Bart and Sam's rows, because our prior clues have now given all the boys possible costumes as : Vampire, Ghoul, or Pirate).

We also know he received "...less than the other boys ", thus we can eliminate (from both Brad's Row, and Vampire column):

Lastly because he(Brad) received "...more than Gina.", we can logically eliminate grid squares :

• This next clue "One of the boys received the most pieces of candy, 2 more than Susie.."

• This clue applies to either Brad, or Sam, thus we can eliminate "ALL" of the girls from column 12:
12 - Gina, Ginger, and Susie.

Now the last part of this clue ("... 2 more than Susie") means :
Susie received 10( or 12 - 2 ), which allows us to clear Susie's row as follows:
Susie - 2, 4, 6, 8, (and consequently) 10's column :

**NOTE** (We can also eliminate Pirate - 10 and Ghoul- 10, because these are Boys costumes, and hence, NOT Susie's).

• This next clue "The Nurse (Ginger) received twice as much candy as the Vampire."

• Provides a solution ( Ginger-Nurse) and some eliminations:
Ginger - Witch, Princess, as well as, Nurse - Gina, Susie, 10, and 12.

Which means the ONLY solution for the child who received 10 pieces must be the Princess, therefore a solution is
Susie-Princess-10 pieces. Let's proceed with the eliminations:

Princess - 2, 4, 6, 8, 12, and Gina.

(Which means the child in the Witch costume must be Gina.

The second part of this clue("...twice as much...as the Vampire") is a little bit trickier:

Because the Vampire (Brad), could only have received 4, 6 or 8 pieces, then Ginger could only have received 8,12, or 16.

(But there is no child who received 16 pieces), therefore we can eliminate
Brad - 8 and Ginger - 2, 4, 6, and 10, as well as ,
Nurse - 2, 4, 6, 10. Which leaves only one solution here : Ginger-Nurse-8,
which in turn yields the solution Brad-4 and the eliminations:

For Column 4 and Column 8:
4 - Bart, Gina, and Sam, as well as, 8 - Bart, Gina, Sam, and for Row 8:
8 - Pirate, Vampire, Ghoul, and Witch.
Because Brad is the Vampire we can eliminate from the Vampire column:
Vampire - 6 and from Row 4 - Ghoul, and Witch.

• This final clue is "Sam received one-half as much as Bart (who was not dressed as a princess or a ghoul)."

• Closes out our Boy's costumes mystery (since we now know Bart is the pirate (because "Bart, not dressed ..ghoul" is not the Ghoul),
meaning Sam must be the Ghoul).
So Sam , the Ghoul received ("...1/2 as much as Bart..."), and the only remaining pieces are 2, 6, 12, then Sam must have received 6, (which is 12/6), Bart 12, and this leaves only 2 for Gina.

• Congratulations! Puzzle solved. To summarize:
Bart-Pirate-12.
Gina-Witch-2.
Ginger-Nurse-8 .
Susie-Princess-10.
Sam-Ghoul-6.
###### PDF PRINTABLE VERSION

4.

THE RACEHORSES-SOLUTION

## THE RACEHORSES- show / hide-

The order of finish:

1.Giddy-up Thom-Grey.
2.Speedy- Midi-White.
3.Diamond- Paul- Silver.
4.Fanny- Shorty- Spotted.
5.Laughy- Matty- Brown.

#### step-by-step:( show / hide)

• This first clue "Matty's horse(Laughy) finished 5th and was not Black."

• Gives us a few good clues, beginning with Matty-Laughy-5th and ending with "....not black".
Locate the following grid squares: [Laughy-5th, Laughy-Matty, Matty-5th] and highlight ALL with 'green boxes'.
Next, make all appropriate grid eliminations in the Laughy row including:
Laughy- 1,2,3,4, and 6.
Laughy- Shorty,Thom, Madge, Midi, and Paul.
Laughy- Black.

Then the grid eliminations for Column 5:
5-Diamond, Fanny, Giddy up, Speedy, Willow, Black, Shorty,Thom, Madge, Midi, and Paul.

The row Matty-1,2,3,4, and 6, as well as the column, Matty-Diamond, Fanny, Giddy up, Speedy, Willow, Black.

• This next clue "Paul's horse (Diamond) was not a winner."

• Locate the following grid squares: [Diamond-1, Paul-1] and make the eliminations for these grid squares.
Next highlight (with a 'green box') the grid square [Diamond-Paul].

Then make the grid eliminations for Row Diamond:

Then make the grid eliminations for Column Paul:
Paul- Fanny, Giddy up, Speedy, and Willow.

• This clue "The horse named Fanny, was the one with the speckles or spots."

• Locate the following grid square: [Fanny-Spotted] and highlight with a 'green box'.

Next, make all appropriate grid eliminations in the Fanny row including:
Fanny - White, Brown, Black, Grey and Silver.

Then the grid eliminations for Column Spotted:
Spotted - Diamond, Giddy up, Laughy,Speedy, and Willow.

**NOTE** Since neither Diamond (owned by Paul) or Laughy (owned by Matty) are the spotted horses, it follows
that: neither Matty or Paul own the spotted horse , and the spotted horse (which is Fran, and which did not
finish 5th
), we can now make these eliminations: [Spotted-5, Spotted-Matty, Spotted-Paul ].

• The next clue "Giddy-up, Speedy, and Diamond finished, (in some order) in the top three as were the following owners Thom, Midi and Paul, whose horses had these characteristics: Grey, White, and Silver,
( also, in some order)."

• This is vital because it tells us three components of the three top finishers, namely:
Horses names: Giddy-up, Speedy, and Diamond; as well as ,
Owners: Thom, Midi and Paul, and the
Colors: Grey, White, and Silver.

(Which allows us to make a multitude of logical conclusions/combinations). To begin with :

Locate the rows for Giddy-up, Speedy, and Diamond, and make the eliminations:

Diamond - 4, 6, (from which this elimination follows Paul - 4, 6).
Giddy Up - 4, 6, Shorty, Madge, Brown, Black.
Speedy - 4,6, Shorty, Madge, Brown, Black.

Next, Locate the rows for Grey, White, and Silver, and make the eliminations:

White - 4, 5, 6, Shorty, and Madge. .
Grey - 4, 5, 6, Shorty, and Madge.
Silver - 4, 5, 6, Shorty, and Madge.

Finally locate these owner's rows: Thom, Midi and Paul (and make the eliminations):
Thom - 4, 6.
Midi - 4, 6.
Paul - 4, 6.

**NOTE** It is important that we now look at the bottom finishers(4th, 5th and 6th), that is these

Horses names: Fanny, Laughy, and Willow; as well as these
Owners: Shorty, Madge, and Matty. and the
Colors: Spotted, Brown, and Black.

(From these combinations we can make some additional eliminations)

First locate the rows for the horses Fanny, Laughy, and Willow and make the eliminations

Fanny - 1,2,3, Thom, Midi,
Laughy - White, Grey, Silver.
Willow - 1,2,3, Thom, Midi, White, Grey, Silver.

Now locate the rows for the non-winning color combos Spotted, Brown, Black (and eliminate)

Spotted - 1, 2, 3, Thom, Midi,
Brown - 1, 2, 3, Thom, Midi,Paul.
Black - 1, 2, 3, Thom, Midi, Paul.

Finally locate the rows for the non-winning owners Shorty, Madge, and Matty and eliminate these combos:

Shorty - 1, 2, and 3.
Madge - 1, 2, and 3.

**NOTE** Before we proceed to the next clue we can make some observations from the grid as follows:

Locate the solution for grid square [Laughy-Brown], (which is the only possible solution remaining for Laughy's row),
and leads to the eliminations in Brown's column :
Brown - Diamond, Speedy, Willow. (Which leads to another solution Willow-Black), and further eliminations of the single grid square [Diamond-Black ],and yet two other solutions [Brown-5th, Brown-Matty ] ,which, in turn ,
allows more eliminations in the Brown row:
Brown - 6, Shorty, and Madge , (Which allows us to complete Matty's column):(by eliminating)

Matty - White, Spotted.

• The next clue is "Speedy the white horse owned by Midi, finished , either, 1st or 2nd."

• We can make an immediate connection namely Speedy-White-Midi , so highlight (with a 'green-box')
The following grid squares in Speedy's row :

[Speedy-Midi, Speedy-White ] while eliminating these

Speedy - Thom, Grey, Silver, and 3 (because, we know, from the end of this clue , that the horse
Speedy, "...finished, either, 1st or second").

**NOTE** ( We can now make some additional eliminations for both Midi, and the White (rows/columns)).

Midi - Giddy Up, Grey, Silver (after highlighting the grid square [Midi-White]), which, of course allows us
to eliminate items in both White's column (White-Diamond , Giddy Up) and row: (White-3, Thom and Paul.).

Lastly, we can eliminate (from Midi's row) the grid square: [Midi-3].

• Lets look at another given clue "Shorty owned either the Spotted horse or , Willow, the one that finished last."

• The only useful portion of this clue (for now) is the last part "...Willow, the one that finished last."

So locate and highlight grid square [Willow-6], then make the appropriate logical eliminations
in Willow's row : Willow - 4,
(which also leads to the solution of Fanny-4) , (after eliminating the grid square [Fanny-6]).

• Now this clue : "Madge does not own the horse that finished 4th. ."

• Let's first locate grid square [Madge-4], make the elimination ,and we discover that Madge
must(and can only) own the horse that finished 6th (- which is Willow, the black horse), therefore, we can finish column 6 eliminations as follows:

6 - Spotted , Shorty (after highlighting 6 - Black, Madge), leaving only Spotted - 4 as the possible solution,
(and after completing the Spotted row), we conclude that Shorty was it's owner , this leaves the final horse-owner
solution (Giddy Up-Thom).

**NOTE** (You can also now complete the 4-Column).

It then remains to determine only the final order of finish!

• Lets look at that last clue "The winning horse did not have a white or silver tail."

• If we start with the elimination of grid squares [White-1, Silver-1], this leaves the "ONLY" solution for the winning horse's color to be'Grey' (so highlight Grey-1), and we may conclude that:

the 2nd place horse had to have been White , and (by consequence), the silver horse was 3rd ,
(and since the white horse finished second), this means

the 2nd place horse could have only have
been Speedy (owned by Midi), which means Diamond (owned by Paul), finished 3rd,
(which was the Silver horse), which now makes the winning horse Giddy Up, which can only be owned by Thom, which has the Grey tail! .

• Congratulations! Puzzle solved. To summarize:
The order of finish:

1.Giddy-up Thom-Grey.
2.Speedy- Midi-White.
3.Diamond- Paul- Silver.
4.Fanny- Shorty- Spotted.
5.Laughy- Matty- Brown.
###### PDF PRINTABLE VERSION

5.

THE CHORES OF ELM STREET-SOLUTION

## CHORES /ELM STREET - show / hide-

Carla-Wed-Mark-Market
Fran-Mon-Tony-Corn.
Nancy-Fri-Stan-Mend.

#### step-by-step:( show / hide)

• Lets look at the first clue "No spouses names begin with the same first letter ."

• Lets look at the next clue "Adriana is not married to either Stan, or Tony, and she did not assign her husband the chore of mending fences."

• Clue "Maxine is married to Adam ,who gets the Eggs on Thursdays."
• This will allow us to add 'green boxes' to the following grid squares:

which allows, in turn, a slew of eliminations, including :
(1)'ALL OTHER squares' in Maxine's row, (2) 'All OTHER squares' in THU Column, (3)'All other squares' in the EGGS row, (4) 'All other squares' in Adam's row, and Adam's column, and finally, (5) 'All other squares' in the Egg's column.

• Our next clue is "Carla is married to either Mark or Bob and sends her husband to Market before Nancy's spouse is assisned his chore ."

• "Carla is married to..." logically implies that she is not married to either Stan, Tony, therefore eliminate

grid squares [Carla-Stan, Carla-Tony], and sends her husband(either Bob , or Mark) to "MARKET" ; so go ahead and add a green box to [Carla-Market] and eliminate "ALL OTHER" squares in the Market Column.

The last part of the clue (.."before Nancy's....), further implies Carla's husband did not perform a chore on Friday, (go ahead and eliminate grid squares [Carla-Friday, Market-Friday]), which also eliminates Nancy from Monday, since no chore can be completed prior to that day(eliminate [Nancy-Mon]).

• Our next clue is "Frans husband's chore is performed on Monday, she is not married to the one who buys the Milk ."

• Start by selecting a 'green box ' for the grid [Fran-Mon], and eliminate 'all other' days for Fran(tue, wed, and fri) as well as, grid squares: [Adriana-Mon, Carla-Mon, Nancy-Mon].
Next we can also eliminate [Fran-Milk, Milk-Mon, Market-Mon] because "..she is not married to the one ....buys milk"

• "Bob's chore is performed earlier in the week than at least 3 others, while Stan mends the fences later in the week than 2 other."

• Bob's clue means he cannot perform chores on Thursday or Friday, so eliminate grid squares: [Bob-Thu, Bob-Fri, Bob-Mend].
Stan by comparision, cannot perform his mending chore on either Monday or Tuesday, therefore eliminate these grid squares: [Stan-Mon, Stan-Tue, Mend-Mon, Mend-Tue]
NOTE: Now some other logical eliminations arise because we can see that:
Fran's chore was Monday, therefore Fran's spouse cannot be Stan, or the one who mends fences, (eliminate these grid squares now : [Fran-Stan, Fran-Mend, ], which , in turn , reveals some other grid square clues: ,[Corn-Mon, Fran-Corn, ] - (each receiving a 'green box') and leads to yet more eliminations: [Adriana-Corn, Carla-Corn, Nancy-Corn], as well as [Corn-Tue, Corn-Wed, Corn-Fri, Corn-Stan],
which allows us to complete the chore list: [Nancy-Mend, Adriana-Milk].

• "No one one is assigned to go to the Market on Tuesday."

• Not much to do here except make the grid square elimination: [Market-Tue].
But wait, this also reveals the only possible remaining day for Market,( Wednesday), which reveals grid squaresto be filed in with 'green boxes' as [Mend-Fri, Milk-Tue], (which in turn yields the following logical conclusions):
Carla-Wed-Market,
Nancy-Thu-mend.

• Our last clue is "Either Stan is married to Fran or Bob buys milk on tuesdays."

• To begin with we have already eliminated Stan-Fran, thus the only true statement here is "Bob buys Milk on Tuesdays",
which since it is true, then Bob can only be the spouse of Adriana, and Bob purchases the Milk.
This leads to Carla's spouse being Mark (which can only mean Mark is the one who goes to Market),which can only mean: Stan must be mending fences on Friday (and therefore, Stan must be the spouse of Nancy),which leaves the only spouse available for Fran , (that being Tony), who since he is Fran's spouse, must be the one who gathers the Corn on Monday.

• Congratulations! Puzzle solved. To summarize:

Adriana's spouse, Bob's chore is performed on Tuesday is to buy Milk.
Carla's spouse, Mark's chore is performed on Wednesday is to go to Market.
Fran's spouse ,Tony's chore is performed on Monday is to buy Corn.
Nancy's spouse, Stan's chore is performed on Friday is to Mend fences.
Maxine's spouse, Adam's chore is performed on Thursday ,is to buy Eggs.
###### PDF PRINTABLE VERSION

6.

TREASURE COVE-SOLUTION

## TREASURE COVE-SOLUTION (show / hide)

Jack - Aqua - Silver.
Green Beard - Dark Blue - Pearls.
One_Eyed Mary - Green - Bronze.
Tortuga - Red - Gold.
Sad Sack - Gold - Map.

#### step-by-step:( show / hide)

• The first clue " The captain of the Green ship did not find the Map ( or the Pearls ) and was not Green Beard ( who did not find the Silver or Bronze ) ."

• Straightforward elimination clue in both Column Green ( ship ) - Map, and Pearls. Then for the Row Green Beard - Green, Silver, Bronze.

• The next clue is " Jack was not the captain of either the Red or Green Ship."

• Again a simple elimination clue , in Row Jack - Red, and Green.

• The next clue states " The Gold ship was piloted by Sad Sack and did not find the Pearls."

• Our first solution clue, locate grid square [Sad Sack - Gold] and click until a 'green box' appears, then proceed with Row eliminations
Sad Sack - Red, Dark Blue, Green, Aqua , and Pearls.

Next we make our Column eliminations : Gold - Jack, Green Beard, One-Eyed Mary, Tortuga and Pearls.

• The next clue " Either Tortuga or Green Beard found the Gold ( but, not in the Gold ship )."

• Provides further elimination clues in both Gold Columns Gold - Jack, One-Eyed Mary, and Sad Sack
and do not forget to eliminate Gold - Gold .

• Our next clue " The one in the Aqua colored ship ( who is not Mary or Green Beard ) found the Silver."

• Begin with these eliminations : Column Aqua - Mary, and Green Beard , then we are given another solution Aqua - Silver,

From which we can now make the eliminations for Row Silver - Red, Dark Blue, Green , and Gold , and in Column Aqua - Pearls, Gold, Map, and Bronze .

• Our last clue is " Either ( Tortuga or Mary ) in the red ship found the Gold."

• On reading this clue, it should come to mind that a prior clue ( " ... Either Tortuga or Green Beard found the Gold..." ), means that Tortuga must be the captain of the red ship ...that found the Gold , which
leads to the following solutions (highlight with 'green boxes'), for Row : Tortuga - Red and Tortuga - Gold , and Column Red - Gold ,

from which we can now make our logical eliminations, back to

Row Tortuga - Dark Blue, Green, Aqua, Silver, Pearls, Map , and Bronze , and Columns Gold - Green Beard, and Red - Green Beard, One-Eyed Mary, Pearls, Map, and Bronze, ( and back to) Row Gold - Dark Blue, Green.

** NOTE ** : ( This will lead to any number of solutions by viewing the different grid square combinations : )

(i) Pearls - Dark Blue which eliminates Dark Blue - Map and Bronze , and

(ii) Map - Gold and eliminates Bronze - Gold , allowing us to solve the final lower combination :

(iii) Bronze - Green .

We also see there remains in the Aqua Column one possibile solution : Aqua - Jack , which by logial deduction makes another solution :

Jack - Silver (in the Row Jack), which allows the following eliminations (still in Row Jack)

Jack - Dark Blue, Pearls, Map, Bronze . And any number of subsequent solutions:

Green Beard - Dark Blue , from which it follows Green Beard - Pearls , and the last solutions:

One-Eyed Mary - Green , One-Eyed Mary - Bronze , and the completion of our game :

• Congratulations! Puzzle solved. To summarize:
Jack - Aqua - Silver.
Green Beard - Dark Blue - Pearls.
One_Eyed Mary - Green - Bronze.
Tortuga - Red - Gold.
Sad Sack - Gold - Map.
###### PDF PRINTABLE VERSION

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